At time t=0, terminal A in the circuit shown in the figure is connected to B by a key and an alternating current I(t)=I0cos(ωt), with I0=1 A and ω=500rads−1 starts flowing in it with the initial direction shown in the figure. At t=6ω7π, the key is switched from B to D. Now onwards only A and D are connected. A total charge Q flows from the battery to charge the capacitor fully. If C=20μF,R=10Ω and the battery is ideal with emf of 50V, identify the correct statement (s).
(A)Magnitude of the maximum charge on the capacitor before t=6ω7π is 1×10−3C.
(B)The current in the left part of the circuit just before t=6ω7π is clockwise.
(C)Immediately after A is connected to D, the current in R is 10A.✓ Correct
(D)Q=2×10−3C.✓ Correct
Answer: (C), (D)
Solution
The current I(t)=I0cos(ωt) charges the capacitor. Integrating: q(t)=ωI0sin(ωt). Maximum charge is 5001=2×10−3 C, not 1×10−3 C, so (A) is wrong.
At t=6ω7π, q=ωI0sin(67π)=−2ωI0=−1×10−3 C (upper plate negative). Current direction at this instant is such that it's anti-clockwise, not clockwise, so (B) is wrong.
When switched to D, the capacitor has −1×10−3 C on the upper plate. Fully charged by 50 V battery: final charge =50×20×10−6=1×10−3 C. Total charge flow Q=1×10−3−(−1×10−3)=2×10−3 C. So (D) is correct.
Immediately after switching, voltage across capacitor is 20×10−6−1×10−3=−50 V. Total voltage across resistor is 50−(−50)=100 V, so current =10100=10 A. So (C) is correct.
JEE Advanced · 2014 · 25 May 2014 · Paper 1 · One or more correct options
